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n-dimensional spheres (Posted on 2014-09-15) Difficulty: 3 of 5

  
An n-dimensional sphere of radius one is intersected with another
n-dimensional sphere of radius one and whose center lies on the
first sphere. The intersection is a (n-1)-dimensional sphere.

The (n-1)-dimensional sphere is intersected with another
n-dimensional sphere of radius one and whose center lies on the
(n-1)-dimensional sphere. The intersection is a (n-2)-dimensional sphere.

This procedure is continued until we have a 1-dimensional sphere.
What is its radius?

This all takes place in an n-dimensional space.
  

See The Solution Submitted by Bractals    
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Comments: ( Back to comment list | You must be logged in to post comments.)
re: The sequence, if my trig is right Comment 2 of 2 |
(In reply to The sequence, if my trig is right by Charlie)

I got the same recurrence relation that you did.

r_k-1^2 = 1 - 1/(4*r_k^2)

But, it seems in a different way.

It took me a little while to get a closed form
for r_k = f(n,k) such that f(n,n) = 1.

r_1 = f(n,1).

Your value for r at n-40 agrees.



  Posted by Bractals on 2014-09-15 09:57:27
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